I am looking for efficient numerical approaches for pricing American options when two or more sources of noise are involved (the simplest case coming to mind would be the Heston Model)

Eventhough I am familiar with lattice methods I don't see how this could work in a "poly-noise" setting. A solution might present itself in the form of MCLSQ (Monte Carlo Least Squares). To my knowldge this method however produces somewhat large deviations ?


  1. What are the prevaliant approaches to dealing with this type of situation numerically ?
  2. How does one benchmark these approximating algorithms if no close form solution for the option-price exists ?
  • $\begingroup$ 1. What's wrong with using a bi/tri-nomial tree to value an american option? Are the instruments you value or the price process more complicated? 2. Are you assuming Least Squares Monte Carlo? If so, it may be worth mentioning in the question. $\endgroup$
    – Oblomov
    Commented Apr 4, 2014 at 9:46
  • $\begingroup$ as far as I know LSQM can produce quite large deviations ? Also if the derivative has multiple sources of noise a lattice approach will be hard to implement. But you are correct I should add some more "meat" to the question $\endgroup$
    – Quanti
    Commented Apr 4, 2014 at 9:48

2 Answers 2


Generally speaking, if you have two or three sources of noise, you are still going to be much better off pricing American options on a lattice than via LSMC. Too often, LSMC becomes the refuge of academics lacking patience to learn proper lattice techniques.

Now, you can frequently reduce the difficulty of pricing American options by considering the american exercise premium $P$, defined as the difference in value between an american-exercise option and its european-exercise equivalent

$$ P = A - E $$

If you have some complicated stochastic model, but enjoy a technique $f(\cdot)$ for pricing european-exercise options

$$ \tilde{E} = f(x_E;\vec\mu) $$

and you can define some much simpler model $g(\cdot)$ that is good enough for estimating the premium

$$ \tilde{P} \approx g(x_A; \vec\nu) - g(x_E; \vec\nu) $$

then your american option price can be estimated as

$$ \tilde{A} \approx \tilde{E} + \tilde{P} $$

If the american exercise premium is large then relative error in $\tilde{P}$ will be important and this trick will not work as well.

Also, if exercise probability is large, or exercise is likely to happen long before the option tenor, then the trick will fail, since we have introduced a dependency on $\vec\mu$ at (european) timescales well past the relevant timescales for the actual american option.

  • $\begingroup$ this is such a great answer - really !! - Brian thank you. Could you add some references on lattice approaches with multiple sources of noise ? $\endgroup$
    – Quanti
    Commented Apr 4, 2014 at 13:16
  • $\begingroup$ as an afterthought: are there any papers comparing LSMC with Lattice Methods $\endgroup$
    – Quanti
    Commented Apr 4, 2014 at 13:37
  • $\begingroup$ I consider Tavella and Randall to be the best place to learn finite difference techniques for finance. $\endgroup$
    – Brian B
    Commented Apr 4, 2014 at 14:30
  • $\begingroup$ @BrianB great reference thank you a lot :) ! $\endgroup$ Commented Apr 5, 2014 at 7:48

Regarding your second question: one possible approach is to reduce the instrument you are trying to value to something simpler, for which an analytical solution are an alternative methodology does exist. You can then vary parameters and check that the valuation is behaving as expected.

If you are using simulations because your price process is more complicated, you can generally tweak the parameters so that it reduces to some other process, which allows an alternative valuation method.

The answer is a bit generic, but your question does not have a lot of detail either.

  • $\begingroup$ Couldy you give some examples of the approaches you menetioned. This is actually exactly the type of information I was looking for !! $\endgroup$
    – Quanti
    Commented Apr 4, 2014 at 11:05
  • $\begingroup$ I also took your suggestions seriously and altered the question accordingly :) $\endgroup$
    – Quanti
    Commented Apr 4, 2014 at 11:19

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